The figure below shows five long parallel wires in the xy plane. Each wire carri

Question

The figure below shows five long parallel wires in the xy plane. Each wire carries a current i = 4.00 A in the positive x direction. The separation between adjacent wires is d = 4.65 cm. What is the net magnetic force on one meter of each wire, due to the other wires? (Enter a numeric answer with N for newtons, and i, j, and k for i hat, j and k as necessary. Note that WebAssign symbolic entry does not accept scientific notation using capital "E", but does accept lower case "e" as well as "*10^".)

Explanation / Answer

a)
It will be attracted by all the other wires
So Fnet will act in +y direction

Fnet = miuo*i*i /(2*pi*d) * (1 + 1/2 + 1/3 + 1/4)
= 2.083*miuo*i*i /(2*pi*d)
= 2.083*(1.2566*10^-6)*4*4/(2*pi*0.0465)
= 1.43*10^-4 N

Answer: 1.43*10^-4 j N

b)
Force due to 1 and 3 will cancel each other
Fnet = force due to 4 and 5
= miuo*i*i /(2*pi*d) * ( 1/2 + 1/3)
= 2.083*miuo*i*i /(2*pi*d)
= 0.833*(1.2566*10^-6)*4*4/(2*pi*0.0465)
= 5.72*10^-5 N
Answer: 5.72*10^-5 jN

c)
Force on 3 will be 0
Answer: 0 N

d)
Force here will be equal and opposite to that on 2
Answer: -5.72*10^-5 j N

e)
Force here will be equal and opposite to that on 1
Answer: -1.43*10^-4 j N

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